Question

$\text{Demonstrati~ca~nr. } \sqrt{n+ \sqrt{n+1} }\text{ este irational, oricare ar fi }n\in \mathbb{N\ast}$

Answer 1

$Pentru~n \in \{1;2 \}~numarul~este~irational. \\ \\ Pentru~n \geq 3~presupunem~prin~reducere~la~absurd~ca~exista~n~a.i. \\ \\ \sqrt{n+ \sqrt{n+1}} \in Q. \\ \\ Daca~ \sqrt{n+ \sqrt{n+1}} \in Q,~atucni~ n+ \sqrt{n+1} \in Q \Rightarrow \sqrt{n+1} \in Q. \\ \\ Proprietate:~Daca~ m \in N~si~ \sqrt{m} \in Q,~atunci~ \sqrt{m} \in N. \\ \\ Conform~acestei~proprietati,~avem~ \sqrt{n+1}=k \in N~(si~k \geq 2) \Rightarrow \\ \\n=k^2-1.~Deci~n+ \sqrt{n+1}=k^2+k-1.$

$Dar~k \geq 2,~de~unde~rezulta~k^2\ \textless \ k^2+k-1\ \textless \ (k-1)^2,~ceea~ce~ \\ \\ inseamna~ca~ \sqrt{k^2+k+1} \notin Q \Leftrightarrow \sqrt{n+ \sqrt{n+1}} \notin Q ~ \forall ~n \in N ,~n \geq 3. \\ \\ Si~cum~ \sqrt{n+ \sqrt{n+1}} \notin Q~pentru~n \in\{1;2\},~rezulta~ca~ \\ \\ \sqrt{n+\sqrt{n+1}} \notin Q ~\forall~n \in N^*.$